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joja [24]
3 years ago
6

Industrial ethanol (CH3CH2OH) is produced by a catalytic reaction of ethylene (CH2═CH2) with water at high pressures and tempera

tures. Calculate ΔH o rxn for this gas-phase hydration of ethylene to ethanol, using bond energies and then using enthalpies of formation.
Chemistry
1 answer:
Natasha2012 [34]3 years ago
6 0

Answer:

using bond energies = - 37 kJ

using enthalpies of formation = - 45.7 kJ

Explanation:

From the reaction;

CH2═CH2(g) + H2O(g) -------> CH3CH2OH(g)

From the above reaction; there are 4 (C-H) bonds , 1 ( C=C) and 2 (H-O) of ethene which forms 5(C-H) bonds, 1 (C-C) and 1 (C-O) and 1(H-O) bonds.

Using Bond Energies; the heat of the reaction can be written as:

\delta H^0_ {rxn}= ∑ energy of old bond breaking + ∑ energies of the new bond formation.

\delta H^0_ {rxn}= ∑ [(4 * BE_{C-H})  + BE_{C-C}) +  + BE_{O-H})]+ ∑ [(5 * BE_{C-H}) + BE_{C-C}) + BE_{O-H}+ BE_{C-O})]

\delta H^0_ {rxn}= ∑[4*413 kJ)+(614kJ)+(2*647kJ]+ ∑ (5*-413kJ)+(-347kJ)+(-467kJ)+(-358kJ)]

\delta H^0_ {rxn}=  -37 kJ

To calculate the heats of reaction by using enthalpies of formation; we have:

\delta H^0_ {rxn}= ∑ \delta H^0_ {products} -  ∑ \delta H^0_ {reactants}

\delta H^0_ {rxn}= ∑ \delta H^0_ {CH_3CH_2OH} - ∑ \delta H^0_ {CH_2=CH_2+H_2O}

\delta H^0_ {rxn}=  (-235.1 kJ) - [(+52.47 kJ) + (-241.826 kJ)]

\delta H^0_ {rxn}=  -45.7 kJ

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Answer:

All are correct

Explanation:

1) The angular momentum quantum number, l, are the subshells within a shell (principle quantum number) it talks about the "form" of an orbital, the number itself tells you about the number of angular nodes (a plane without electronic density). It starts at l=0 where you don't see any nodes and it takes the form of an sphere, and we knowing it bu another name an s-orbital. It takes values up to n-1.

l=0 (sphere - s-orbital)

l=1 (p-orbital)

l=2 (d-orbital)

2) The magnetic quatum number, ml relates to the number of orbitals within a subshell then it is related with l, taking values form -l to l incluing 0.

For l=0 (s-orbital) ml=0

For l=1 (p-orbital) ml=1,0,-1

For l=2 (d-orbital) ml=2,1,0,-1,-2

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8 0
3 years ago
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Answer:

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